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Ta có:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)
\(=1-\frac{2n+1}{\left(n+1\right)^2}\)
Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
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NHỚ ĐỌC KỸ ĐỀ ĐẤY
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\(x\left(x+2\right)\left(x^2+2x+2\right)+1\)
\(=\left(x^2+2x\right)\left(x^2+2x+2\right)+1\)
Đặt: \(x^2+2x=t\)
khi đó: \(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=t\left(t+2\right)+1=\left(t+1\right)^2\)
\(=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)
b) Xét: \(\left(n+1\right)^2-n^2=\left(n+1+n\right)\left(n+1-n\right)=2n+1\)
Khi đó:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)
\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
1,tổng quát: (2k+1)/[k(k+1)^2]
=(2k+1)/k^2(k+1)^2=[(k+1)^^2-k^2]/k^2(k+1)^2=1/k^2-1/(k+1)^2
áp dụng vào ,kết quả =2024/2025
\(A=\left(\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}+\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}\right):\dfrac{x-1}{x^3}\)
\(=\dfrac{x^2+3}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)^2}\)